If one earthquake has a magnitude of $(5.2)$ on the Richter scale and second earthquake has a magnitude of $(6)$ on the Richter scale. Compare the intensities of two earthquakes.

$$
L=10logleft(dfrac{I}{I_{o}}right) 5.2=10logleft(dfrac{I}{I_{o}}right)
$$

$$
L=10logleft(dfrac{I}{I_{o}}right) 6=10logleft(dfrac{I}{I_{o}}right)
$$

Divide both of sides by $(10)$ as follows:

$$
0.52=logleft(dfrac{I}{I_{o}}right)
$$

$$
0.6=logleft(dfrac{I}{I_{o}}right)
$$

Use the exponential standard form as follows:

$$
because log_{10}(x)=b x=10^b
$$

$$
therefore left(dfrac{I_{5.2}}{I_{o}}right)=10^{0.52}
$$

$$
therefore left(dfrac{I_{6}}{I_{o}}right)=10^{0.6}
$$

Isolate the $(I)$ in the left side as follows:

$$
I_{5.2}=10^{0.52}cdot I_{o}
$$

$$
I_{6}=10^{0.6}cdot I_{o}
$$

$$
dfrac{I_{5.2}}{I_{6}}=dfrac{10^{0.52}cdot I_{o}}{10^{0.6}cdot I_{o}} dfrac{I_{5.2}}{I_{6}}=10^{-0.08}=dfrac{1}{10^{0.08}}=0.83176
$$

$$
text{color{#4257b2}$therefore $The sound of the first earthquake is $(0.83176)$ times less intense than the second earthquake.}
$$

$$
text{color{#c34632}The sound of the first earthquake is $(0.83176)$ times less intense than the second earthquake.}
$$

Calculate the PH of swimming pool with a concentration of hydrogen ion with $(6.21times10^{-8})$ Mole/L

According to the standard form of the concentration equation $PH=-log(H^+)$

Substitute the value of hydrogen concentration as follows:

$$
PH=-log(6.21times10^{-8})
$$

$$
PH=-(-7.2069) PH=7.2069
$$

$$
text{color{#c34632}$PH=7.2069$}
$$

The loudness of sound measured in decibels is
$$
L=10logdfrac{I}{I_0},tag{1}$$
where $I$ is the intensity of a sound being measured and $I_0$ is the intensity of a sound at a threshold of a human hearing, or the intensity of sound humans can barely hear.

If the intensity of a sound is $1.000.000$ times more intense than the intensity at a threshold of a human hearing, we can substitute $I=1.000.000 I_0$ in Eq. $(1)$ and evaluate for loudness $L$. It yields:
$$
begin{align*}
L&=10logdfrac{1.000.000cancel{I_0}}{cancel{I_0}}\
&=10log1.000.000&&[1.000.000=10^6]\
&=10log10^6&&[log10^n=nlog10]\
&=6cdot 10 log 10&&[log10=1]\
&=60.end{align*}$$
Hence, the loudness of the sound that is $1.000.000$ times more intense than the threshold sound is $60 mathrm{dB}$.

$60 mathrm{dB}$

$$
text{color{#4257b2}The loudness of a heavy snore is $(69)$ dB. How many times as should as a normal conversation of $(60)$ dB is a heavy one.}
$$

$$
L=10logleft(dfrac{I}{I_{o}}right) 69=10logleft(dfrac{I}{I_{o}}right)
$$

$$
L=10logleft(dfrac{I}{I_{o}}right) 60=10logleft(dfrac{I}{I_{o}}right)
$$

Divide both of sides by $(10)$ as follows:

$$
6.9=logleft(dfrac{I}{I_{o}}right)
$$

$$
6=logleft(dfrac{I}{I_{o}}right)
$$

Use the exponential standard form as follows:

$$
because log_{10}(x)=b x=10^b
$$

$$
therefore left(dfrac{I_{69}}{I_{o}}right)=10^{6.9}
$$

$$
therefore left(dfrac{I_{60}}{I_{o}}right)=10^{6}
$$

Isolate the $(I)$ in the left side as follows:

$$
I_{69}=10^{6.9}cdot I_{o}
$$

$$
I_{60}=10^{6}cdot I_{o}
$$

$$
dfrac{I_{69}}{I_{60}}=dfrac{10^{6.9}cdot I_{o}}{10^{6}cdot I_{o}} dfrac{I_{69}}{I_{60}}=10^{0.9}=7.943
$$

$$
text{color{#4257b2}$therefore$ The sound of the heavy snore is $(7.943)$ times more intense than the normal conversation.}
$$

$$
text{color{#c34632}The sound of the heavy snore is $(7.943)$ times more intense than the normal conversation.}
$$

Calculate the hydrogen ion concentration for the following items.

$$
text{color{#4257b2}(a) Backing soda, with a $PH=9$}
$$

According to the standard form of the concentration equation $PH=-log(H^+)$

$$
9=-log(H^+)
$$

Multiply the entire equation by $(-1)$ as follows:

$$
-9=log(H^+)
$$

Use the standard form for the exponential equation as follows:

$$
because log_a(x)=b therefore x=a^b
$$

$$
because log(H^+)=-9 therefore H^+=10^{-9}
$$

$$
H^+=10^{-9}
$$

$$
text{color{#4257b2}(b) Milka, with a $PH=6.6$}
$$

According to the standard form of the concentration equation $PH=-log(H^+)$

$$
6.6=-log(H^+)
$$

Multiply the entire equation by $(-1)$ as follows:

$$
-6.6=log(H^+)
$$

Use the standard form for the exponential equation as follows:

$$
because log_a(x)=b therefore x=a^b
$$

$$
because log(H^+)=-6.6 therefore H^+=10^{-6.6}
$$

$$
H^+=2.51times10^{-7}
$$

$$
text{color{#4257b2}(c) An egg, with a $PH=7.8$}
$$

According to the standard form of the concentration equation $PH=-log(H^+)$

$$
7.8=-log(H^+)
$$

Multiply the entire equation by $(-1)$ as follows:

$$
-7.8=log(H^+)
$$

Use the standard form for the exponential equation as follows:

$$
because log_a(x)=b therefore x=a^b
$$

$$
because log(H^+)=-7.8 therefore H^+=10^{-7.8}
$$

$$
H^+=1.585times10^{-8}
$$

$$
text{color{#4257b2}(d) Oven cleaner, with a $PH=13$}
$$

According to the standard form of the concentration equation $PH=-log(H^+)$

$$
13=-log(H^+)
$$

Multiply the entire equation by $(-1)$ as follows:

$$
-13=log(H^+)
$$

Use the standard form for the exponential equation as follows:

$$
because log_a(x)=b therefore x=a^b
$$

$$
because log(H^+)=-13 therefore H^+=10^{-13}
$$

$$
H^+=10^{-13}
$$

$$
text{color{#c34632}(a) $H^+=10^{-9}$ (b) $H^+=2.51times10^{-7}$
\ \
(c) $H^+=1.585times10^{-8}$ (d) $H^+=10^{-13}$}
$$

Calculate the PH of the solution for the following items.

$$
text{color{#4257b2}(a) Concentration of $H^+=0.00032$}
$$

According to the standard form of the concentration equation $PH=-log(H^+)$

$$
PH=-log(0.00032)
$$

$$
PH=-(-3.495) PH=3.495
$$

$$
text{color{#4257b2}(b) Concentration of $H^+=0.0003$}
$$

According to the standard form of the concentration equation $PH=-log(H^+)$

$$
PH=-log(0.0003)
$$

$$
PH=-(-3.523) PH=3.523
$$

$$
text{color{#4257b2}(c) Concentration of $H^+=0.000045$}
$$

According to the standard form of the concentration equation $PH=-log(H^+)$

$$
PH=-log(0.000045)
$$

$$
PH=-(-4.347) PH=4.347
$$

$$
text{color{#4257b2}(d) Concentration of $H^+=0.005$}
$$

According to the standard form of the concentration equation $PH=-log(H^+)$

$$
PH=-log(0.005)
$$

$$
PH=-(-2.301) PH=2.301
$$

$$
text{color{#c34632}(a) $PH=3.495$ (b) $PH=3.523$
\ \
(c) $PH=4.347$ (d) $PH=2.301$}
$$

(a) Calculate the PH of distilled water which has a concentration of

$$
color{#4257b2}(H^+)=10^{-7}
$$

According to the standard form of the concentration equation $PH=-log(H^+)$

$$
PH=-log(10^{-7})
$$

$$
PH=-(-7) PH=7
$$

(b) Drinking water from a particular tap has a PH between $color{#4257b2}6.3, 6.6$.
Is this tap more or less acidic than distilled water?

The PH of distilled water is $PH=7$

Since the PH of water particular tap is less than $(7)$, so this tap is more acidic than the distilled water.

$$
text{color{#c34632}(a) $PH=7$
\ \
(b) Since the PH of water particular tap is less than $(7)$, so this tap is more acidic than the distilled water.}
$$

$$
text{color{#4257b2}The loudness of moving power lawn is $(109)$ dB, the noise level in front of amplifier is a bout $(118)$ dB.\ \ How many times the noise level in front of amplifier than the loudness of moving power lawn .}
$$

$$
L=10logleft(dfrac{I}{I_{o}}right) 118=10logleft(dfrac{I}{I_{o}}right)
$$

$$
L=10logleft(dfrac{I}{I_{o}}right) 109=10logleft(dfrac{I}{I_{o}}right)
$$

Divide both of sides by $(10)$ as follows:

$$
11.8=logleft(dfrac{I}{I_{o}}right)
$$

$$
10.9=logleft(dfrac{I}{I_{o}}right)
$$

Use the exponential standard form as follows:

$$
because log_{10}(x)=b x=10^b
$$

$$
therefore left(dfrac{I_{118}}{I_{o}}right)=10^{11.8}
$$

$$
therefore left(dfrac{I_{109}}{I_{o}}right)=10^{10.9}
$$

Isolate the $(I)$ in the left side as follows:

$$
I_{118}=10^{11.8}cdot I_{o}
$$

$$
I_{109}=10^{10.9}cdot I_{o}
$$

$$
dfrac{I_{118}}{I_{109}}=dfrac{10^{11.8}cdot I_{o}}{10^{10.9}cdot I_{o}} dfrac{I_{118}}{I_{109}}=10^{0.9}=7.943
$$

$$
text{color{#4257b2}$therefore$ The sound of the noise level in front of the amplifier is $(7.943)$ times more intense than the loudness of moving power lawn.}
$$

$$
text{color{#c34632}The sound of the noise level in front of the amplifier is $(7.943)$ times more intense than the loudness of moving power lawn.}
$$

Determine the depth of water, $x$ at which most aquatic plants at receive the required light where,

The intensity, $I$, of a light represent by $I=10^{1-0.13x}$. Most aquatic plants require alight intensity of $4.2$ units of strong growth.

$$
4.2=10^{1-0.13x}
$$

Write the value of $4.2$ as a power of $10$ as follows:

$$
10^{0.624}=10^{1-0.13x}
$$

Since the base of eponential form is equals, so the power is equal as follows:

$$
0.624=1-0.13x
$$

Isolate the variables of $x$ in the left side as follows:

$$
0.13x=1-0.624 0.13x=0.376
$$

Divide both of sides by $(0.13)$ as follows:

$$
x=dfrac{0.376}{0.13} x=2.892
$$

$$
text{color{#4257b2}$therefore$ The depth of water is $x=2.892$ meter}
$$

$$
text{color{#c34632}The depth of water is $x=2.892$ meter}
$$

*a)* The table gives the growth in the number of bacteria over $42$ hours. To obtain the model that best fits this data we can make a scatter plot and find the exponential equation that best fits the given data using the graphing calculator and built-in functions. To obtain the same model algebraically, we can use the exponential equation of a form
$$
y=ab^t,tag{1}$$
where $y$ is the number of bacteria after $t$ hours, $a$ is the initial number of bacteria and $b=1+r$, where $r$ is the growth rate of bacteria.

We can start by entering the given data into the list. By pressing **STAT** then **ENTER** it will open a list where we can input our values. Into the list $L_1$ we should enter the data for hours, pressing **ENTER** after each value to input the next one. Into column $L_2$ we should input the values of the bacteria number, again pressing **ENTER** after each value.

Now, to make a scatter plot, press **2nd**, **Y=**, and **1** **ENTER**. Make sure that the plot is set to *ON* and choose the type and color of the graph you want to display. Press **ZOOM** and **9** to display the graph. It should look as depicted in *Figure 1$.

Figure 1.

To determine the equation that models this situation, i.e. the curve that best fits this data on the graph, we should press **STAT** and scroll to the right over **CALC**. By pressing **0 ExpReg** we will choose the exponential regression since the exponential function of a form $y=ab^x$ best fits the data in the table. By pressing **ENTER** the calculator gives
$$
a=850 text{and} b=1.15$$
which gives the exponential equation of a form
$$ y=850(1.15)^t.tag{2}$$

Now, using Eq. $(1)$ by plugging in the initial value $a=850$, the value after $t=7$ hours $y=2250$ we can solve for $b$ and rewrite that equation in a general form. Hence, we obtain:
$$
begin{align*}
2250&=850b^7&&[text{Divide by $850$ throughout}]\
2.65&=b^7&&[text{ Take $7$th root of both sides and reverse}]\
sqrt[7]{b^7}&=sqrt[7]{2.65}
\
b
&=1.15.end{align*}$$
Hence, the function that best fits the given date is of the form of Eq. $(1)$ and it yields
$$
y=850(1.15)^t.$$
Comparing this equation with the equation $(2)$ we’ve obtained graphically using a calculator, we can conclude that both results are the same.

*b)* If the initial number of bacteria was $a=850$ then the double bacteria number is $y=2cdot850=1700$. Now using these values with the Eq. $(1)$ and setting $b=1.15$ we can calculate the time needed for the bacteria number to double by solving the exponential equation by using basic properties of a logarithm. It yields:
$$
begin{align*}
1700&=850(1.15)^t&&[text{Divide by $850$ throughout}]\
2&=1.15^t&&[text{Apply logarithm of base $10$}]\
log2&=log 1.15^t&&[log a^x=xlog a]\
log2&=tlog 1.15\
t&=frac{log2}{log 1.15}\
&approx4.95 mathrm{hours}.
end{align*}

a) $y=850(1.15)^t$
b) $tapprox4.95 mathrm{hours}$

The function of purifying the cleaning fluid is exponential so that it has a form
$$
A=A_0(1-r)^xtag{1}$$
where $A$ is the amount of fluid after $x$ times of purification, $A_0$ is the initial amount of fluid, and $r$ is the rate of change at which the fluid evaporates.

When only one half of fluid remains we can set $A=frac{A_0}{2}$ into Eq. $(1)$. Furthermore, rate of change $2.1%$ we can rewrite as a decimal $r=frac{2.1}{100}=0.021$ and also substitute in the give equation and solve for $x$. It yields:
$$
begin{align*}
dfrac{cancel{A_0}}{2}&=cancel{A_0}(1-0.021)^x\
0.5&=0.979^x\
x&=frac{log 0.5}{log 0.979}&&[a^x=biff log_ab=x]\
x&approx33 mathrm{times}
end{align*}$$

Hence, after about $33$ cycles of purification, only half the fluid remains and it needs to be topped up.

$approx33 mathrm{times}$

$$
text{color{#4257b2}How long will it take for $(2500)$ dollar to accumulate $(4000)$ dollars if the interest rate is $(6.5)$ hundred percent per year.}
$$

Interest rate per year is $.065cdot2500=162.5$ dollar per year

$a_{0}=2500$ dollar

$$
a_{1}=2500+162.5=2662.5=a_{0}+162.5
$$

$$
a_{2}=2662.5+162.5=2825=a_{1}+162.5=a_{0}+162.5+162.5=a_{0}+2(162.5)
$$

$$
color{#4257b2}a_{n}=2500+162.5n
$$

$$
4000=2500+162.5n
$$

Isolate the variables on the left side as follows:

$$
4000-2500=162.5n 162.5n=1500
$$

Divide both of sides by $(162.5)$ as follows:

$$
n=dfrac{1500}{162.5} n=9.23
$$

The time needed to accumulate $(4000)$ dollars is $(9.23)$ year.

$$
text{color{#c34632}The time needed to accumulate $(4000)$ dollars is $(9.23)$ year.}
$$

$$
text{color{#4257b2}How many days will be $(75)$ hundred percent of the wound be healed for the formula $A(t)=80(10)^{-0.023t}$}
$$

$$
A(t)=80(10)^{0.023t} 0.75=80(10)^{-0.023t}
$$

Use logarithmic formula for both sides as follows:

$$
log(0.75)=logleft(80 (10)^{-0.023t}right)
$$

According the standard form as follows:

$$
log_a(xy)=log_a(x)+log_a(y)
$$

$$
log(0.75)=log(80)+log(10)^{-0.023t}
$$

$$
log(0.75)=log(80)-0.023t log(10)
$$

$$
-0.1249=1.9031-0.023 t
$$

Isolate the variables on the left side as follows:

$$
0.023 t=1.9031+.01249 0.023 t=2.028
$$

Divide both of sides by $(0.023)$ as follows:

$$
t=dfrac{2.028}{0.023} t=88.1739
$$

$$
text{color{#4257b2}The time needed to will be $(75)$ hundred percent of the wound be healed is $(88.1739)$ days.}
$$

$$
text{color{#c34632}The time needed to will be $(75)$ hundred percent of the wound be healed is $(88.1739)$ days.}
$$

$$
text{color{#4257b2}Calculate the sound level in the second car if a new car has a a sound level of $(70)$ dB with speed of $(50)$ Km/hr.}
$$

For the new car

$$
L=10logleft(dfrac{I}{I_{0}}right) 70=10logleft(dfrac{I}{I_{0}}right)
$$

Divide both of sides by $(10)$ as follows:

$$
7=logleft(dfrac{I}{I_{0}}right)
$$

According standard formula, as follows:

$$
because log_a(x)=b therefore x=a^b
$$

$$
10^7=dfrac{I}{I_{0}} I=10^7 I_{0}
$$

For the second car

$$
L=10logleft(dfrac{I}{I_{0}}right) L=10logleft(dfrac{2cdot10^7 I_{0}}{I_{0}}right)
$$

$$
L=10log(2cdot10^7)
$$

$$
L=10cdot7.301
$$

$$
L=73.01text{ dB}
$$

$$
text{color{#4257b2}The sound level in the second car is $L=73.01text{ dB}$}
$$

$$
text{color{#c34632}The sound level in the second car is $L=73.01text{ dB}$}
$$

The given inflation is an exponential function of a form
$$
C=P(1+r)^t,tag{1}$$
where $P$ is the initial cost of some good, $C$ is the cost after $t$ years and $r$ is the rate of inflation as a decimal number.

Using this equation and the given information we can compute the desired values.

*a)* Given that the rate of inflation is $r=3.8%$ we can rewrite it as a decimal number by dividing it by $100$. It gives $r=0.038$ so that the factor $1+r=1.038$. Hence, using Eq. $(1)$ we can compute the current cost $C$ of any good after $t$ years knowing its initial value $P$ as
$$C=P(1.038)^t.$$

*b)* Using the Eq. $(1)$, the growth factor we’ve obtained in the previous step $1+r=1.038$ and initial value of a break $P=$400$, the cost of a break after $t=10$ years is
$$
C=$400cdot1.038^{10}=$580.81.$$

*c)* Using the Eq. $(1)$, the growth factor we’ve obtained in the first step $1+r=1.038$ and value of an oil change $C=$47.95$ in $t=10$ years, the initial value of an oil change is
$$
P=dfrac{C}{(1+r)^t}=dfrac{$47.95}{1.038^{10}}approx$33.02.$$

a) $C=P(1.038)^t$
b) $C=$580.81$
c) $P=$33.02$